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Given the following test scores, find a 95 percent confidence interval for the p...
Oct 3, 2024
Given the following test scores, find a 95 percent confidence interval for the population mean: 148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236. Assume population normality.
Solution by Steps
step 1
First, we calculate the sample mean (xˉ\bar{x}) of the given test scores: xˉ=148+154+158+160+161+162+166+170+182+195+23611 \bar{x} = \frac{148 + 154 + 158 + 160 + 161 + 162 + 166 + 170 + 182 + 195 + 236}{11}
step 2
Next, we find the sample standard deviation (ss) using the formula: s=∑(xi−xˉ)2n−1 s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} , where n n is the sample size (11 in this case)
step 3
We then determine the critical value for a 95% confidence interval using the t-distribution. For n−1=10 n-1 = 10 degrees of freedom, the critical value t0.025 t_{0.025} can be found in t-tables or using statistical software
step 4
Finally, we calculate the confidence interval using the formula: xˉ±t0.025⋅sn \bar{x} \pm t_{0.025} \cdot \frac{s}{\sqrt{n}}
Answer
The 95% confidence interval for the population mean is approximately (164.45, 195.55).
Key Concept
Confidence intervals provide a range of values that likely contain the population mean based on sample data.
Explanation
The calculated confidence interval indicates that we are 95% confident that the true population mean lies within this range.
Solution by Steps
step 1
First, we calculate the sample mean (xˉ\bar{x}) of the given test scores: xˉ=148+154+158+160+161+162+166+170+182+195+23611=168.27 \bar{x} = \frac{148 + 154 + 158 + 160 + 161 + 162 + 166 + 170 + 182 + 195 + 236}{11} = 168.27
step 2
Next, we find the sample standard deviation (ss) using the formula: s=∑(xi−xˉ)2n−1 s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} where n=11n = 11. After calculating, we find s≈27.45s \approx 27.45
step 3
Now, we calculate the standard error (SE) of the mean: SE=sn=27.4511≈8.26 SE = \frac{s}{\sqrt{n}} = \frac{27.45}{\sqrt{11}} \approx 8.26
step 4
Finally, we find the 95% confidence interval using the formula: CI=xˉ±t∗⋅SE CI = \bar{x} \pm t^* \cdot SE where t∗t^* for 10 degrees of freedom at 95% confidence is approximately 2.228. Thus, CI=168.27±2.228⋅8.26 CI = 168.27 \pm 2.228 \cdot 8.26 which gives us the interval approximately (155.24, 181.30)
Answer
A
Key Concept
Confidence intervals provide a range of values that likely contain the population mean based on sample data.
Explanation
The confidence interval is calculated using the sample mean, standard deviation, and the t-distribution to account for sample size.
What sample size is needed to obtain a 90 percent confidence interval for the mean protein content of meat if the estimate is to be within 2 pounds of the true mean value? Assume that the variance is 49 pounds.
Solution by Steps
step 1
To determine the required sample size n n for a 90% confidence interval, we use the formula: n=(Z⋅σE)2 n = \left( \frac{Z \cdot \sigma}{E} \right)^2 where Z Z is the Z-score for 90% confidence, σ \sigma is the population standard deviation, and E E is the margin of error
step 2
For a 90% confidence level, the Z-score is approximately Z=1.645 Z = 1.645 . Given that the variance is 49 pounds, the standard deviation σ \sigma is 49=7 \sqrt{49} = 7 pounds. The margin of error E E is 2 pounds
step 3
Substituting the values into the formula: n=(1.645⋅72)2 n = \left( \frac{1.645 \cdot 7}{2} \right)^2 which simplifies to n=(11.5152)2=(5.7575)2≈33.05 n = \left( \frac{11.515}{2} \right)^2 = (5.7575)^2 \approx 33.05
step 4
Since the sample size must be a whole number, we round up to the nearest whole number, giving us n=34 n = 34
Answer
The required sample size is 34.
Key Concept
Sample size determination for confidence intervals.
Explanation
The sample size is calculated based on the desired confidence level, population standard deviation, and margin of error to ensure the estimate is within the specified range.
In a study of factors affecting whether soldiers decide to reenlist, 320 subjects were measured for an index of satisfaction. The sample mean is 28.8 and the sample standard deviation is 7.3. Use the given sample data to construct the 98 percent confidence interval for the population mean for level of satisfaction.
Solution by Steps
step 1
To construct a 98% confidence interval for the population mean, we first need to find the Z-score corresponding to a 98% confidence level. The Z-score for 98% confidence is approximately 2.33
step 2
Next, we calculate the standard error (SE) using the formula: SE=sn SE = \frac{s}{\sqrt{n}} where s=7.3 s = 7.3 and n=320 n = 320 . Thus, SE=7.3320≈0.41. SE = \frac{7.3}{\sqrt{320}} \approx 0.41.
step 3
Now, we can calculate the margin of error (ME) using the formula: ME=Z⋅SE=2.33⋅0.41≈0.96. ME = Z \cdot SE = 2.33 \cdot 0.41 \approx 0.96.
step 4
Finally, we construct the confidence interval using the sample mean and the margin of error: CI=(xˉ−ME,xˉ+ME)=(28.8−0.96,28.8+0.96)=(27.84,29.76). CI = (\bar{x} - ME, \bar{x} + ME) = (28.8 - 0.96, 28.8 + 0.96) = (27.84, 29.76).
Answer
The 98% confidence interval for the population mean level of satisfaction is approximately (27.84, 29.76).
Key Concept
Confidence intervals provide a range of values that likely contain the population mean based on sample data.
Explanation
The calculated confidence interval indicates that we can be 98% confident that the true mean level of satisfaction for soldiers falls between 27.84 and 29.76.
The coffee and soup machine at the local subway station is supposed to fill cups with 6 ounces of soup. Ten cups of soup are bought with results of a mean of 5.93 ounces and a standard deviation of .13 ounces. How large a sample of soups would we need to be 95 percent confident that the sample mean is within .03 ounces of the population mean?
Solution by Steps
step 1
To determine the required sample size n n for a 95% confidence interval, we use the formula: n=(Z⋅σE)2 n = \left( \frac{Z \cdot \sigma}{E} \right)^2 where Z Z is the Z-score for 95% confidence, σ \sigma is the population standard deviation, and E E is the margin of error
step 2
For a 95% confidence level, the Z-score is approximately 1.96. Given that the standard deviation σ=0.13 \sigma = 0.13 ounces and the margin of error E=0.03 E = 0.03 ounces, we substitute these values into the formula: n=(1.96⋅0.130.03)2 n = \left( \frac{1.96 \cdot 0.13}{0.03} \right)^2
step 3
Calculating the above expression: n=(0.25480.03)2=(8.4933)2≈72.14 n = \left( \frac{0.2548}{0.03} \right)^2 = \left( 8.4933 \right)^2 \approx 72.14 We round up to the nearest whole number, giving us n=73 n = 73
Answer
The required sample size is 73.
Key Concept
Sample size determination for confidence intervals
Explanation
The sample size is calculated to ensure that the sample mean is within a specified margin of error from the population mean at a given confidence level.
A manufacturer of NFL regulation footballs uses a machine to inflate its new balls to a pressure of 13.5 pounds (σ = .1). When the machine is properly calibrated, the mean inflation pressure is 13.5 pounds, but uncontrollable factors can cause the pressure of individual footballs to vary. For quality control purposes, the manufacturer wishes to estimate the mean inflation pressure to within .025 pounds of its true value with 99 percent confidence. What sample size should be used?
Solution by Steps
step 1
To determine the required sample size n n for a 99% confidence interval, we use the formula: n=(Z⋅σE)2 n = \left( \frac{Z \cdot \sigma}{E} \right)^2 where Z Z is the Z-score for 99% confidence, σ=0.1 \sigma = 0.1 pounds, and E=0.025 E = 0.025 pounds
step 2
For a 99% confidence level, the Z-score is approximately 2.576. Thus, we substitute the values into the formula: n=(2.576⋅0.10.025)2 n = \left( \frac{2.576 \cdot 0.1}{0.025} \right)^2
step 3
Calculating the expression gives us: n=(0.25760.025)2=(10.304)2≈106.16 n = \left( \frac{0.2576}{0.025} \right)^2 = \left( 10.304 \right)^2 \approx 106.16
step 4
Since the sample size must be a whole number, we round up to the nearest whole number, giving us n=107 n = 107
Answer
The required sample size is 107.
Key Concept
Sample size determination for confidence intervals.
Explanation
The answer reflects the necessary sample size to estimate the mean inflation pressure with a specified margin of error and confidence level.
A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of .996 ounces with a standard deviation of .004 ounces. How many pieces of candy must we sample if we want to be 99 percent confident that the sample mean is within .001 ounces of the true mean?
Solution by Steps
step 1
To determine the required sample size n n for a 99% confidence interval, we use the formula: n=(Z⋅σE)2 n = \left( \frac{Z \cdot \sigma}{E} \right)^2 where Z Z is the Z-score for 99% confidence, σ=0.004 \sigma = 0.004 ounces, and E=0.001 E = 0.001 ounces
step 2
For a 99% confidence level, the Z-score is approximately 2.576. Thus, we substitute the values into the formula: n=(2.576⋅0.0040.001)2 n = \left( \frac{2.576 \cdot 0.004}{0.001} \right)^2
step 3
Calculating the above expression gives us: n=(0.0103040.001)2=(10.304)2≈106.16 n = \left( \frac{0.010304}{0.001} \right)^2 = (10.304)^2 \approx 106.16
step 4
Since the sample size must be a whole number, we round up to the nearest whole number, giving us n=107 n = 107
Answer
The required sample size is 107.
Key Concept
Sample size determination for confidence intervals.
Explanation
The calculation ensures that the sample mean is within the specified margin of error at the desired confidence level.
A sample of 2,000 people yielded = .52. What is the variance of the population proportion? 答案选项组
Solution by Steps
step 1
To find the variance of the population proportion, we use the formula: σ2=p(1−p)n \sigma^2 = \frac{p(1 - p)}{n} where p=0.52 p = 0.52 and n=2000 n = 2000
step 2
Substituting the values into the formula: σ2=0.52(1−0.52)2000=0.52⋅0.482000 \sigma^2 = \frac{0.52(1 - 0.52)}{2000} = \frac{0.52 \cdot 0.48}{2000}
step 3
Calculating the variance: σ2=0.24962000=0.0001248 \sigma^2 = \frac{0.2496}{2000} = 0.0001248
Answer
The variance of the population proportion is 0.0001248.
Key Concept
The variance of a population proportion measures the variability of the proportion in the population.
Explanation
The variance is calculated using the formula for the population proportion, which accounts for the sample size and the proportion itself.
A sample of 100 items has a population standard deviation of 5.1 and a mean of 21.6. Construct a 95 percent confidence interval for μ.
Solution by Steps
step 1
To construct a 95% confidence interval for the population mean μ \mu , we first need to determine the critical Z-value for a 95% confidence level. This value is approximately Z=1.96 Z = 1.96
step 2
Next, we calculate the standard error (SE) of the mean using the formula: SE=σn SE = \frac{\sigma}{\sqrt{n}} where σ=5.1 \sigma = 5.1 and n=100 n = 100 . Thus, SE=5.1100=5.110=0.51. SE = \frac{5.1}{\sqrt{100}} = \frac{5.1}{10} = 0.51.
step 3
Now, we can calculate the margin of error (ME) using the formula: ME=Z×SE=1.96×0.51≈1.00. ME = Z \times SE = 1.96 \times 0.51 \approx 1.00.
step 4
Finally, we construct the confidence interval using the sample mean xˉ=21.6 \bar{x} = 21.6 : CI=xˉ±ME=21.6±1.00. CI = \bar{x} \pm ME = 21.6 \pm 1.00. Therefore, the confidence interval is (20.6,22.6) (20.6, 22.6)
Answer
The 95% confidence interval for μ \mu is (20.6,22.6) (20.6, 22.6) .
Key Concept
Confidence intervals provide a range of values within which we expect the population mean to lie with a certain level of confidence.
Explanation
The calculated confidence interval indicates that we are 95% confident that the true population mean falls between 20.6 and 22.6.
Solution by Steps
step 1
First, we need to determine the critical value for a 98% confidence interval. Since the sample size n=15 n = 15 , we will use the t-distribution. The degrees of freedom df=n−1=14 df = n - 1 = 14 . The critical t-value for 98% confidence can be found using a t-table or calculator: t0.01,14≈2.624 t_{0.01, 14} \approx 2.624
step 2
Next, we calculate the standard error (SE) of the sample mean using the formula: SE=sn=0.615≈0.155 SE = \frac{s}{\sqrt{n}} = \frac{0.6}{\sqrt{15}} \approx 0.155
step 3
Now, we can calculate the margin of error (ME) using the formula: ME=t×SE=2.624×0.155≈0.407 ME = t \times SE = 2.624 \times 0.155 \approx 0.407
step 4
Finally, we can construct the confidence interval using the sample mean Xˉ=1.2 \bar{X} = 1.2 : CI=(Xˉ−ME,Xˉ+ME)=(1.2−0.407,1.2+0.407)=(0.793,1.607) CI = \left( \bar{X} - ME, \bar{X} + ME \right) = (1.2 - 0.407, 1.2 + 0.407) = (0.793, 1.607)
Answer
A
Key Concept
Confidence intervals provide a range of values that likely contain the population mean based on sample data.
Explanation
The confidence interval is calculated using the sample mean, the critical value from the t-distribution, and the standard error, which accounts for sample size and variability.
Solution by Steps
step 1
The width of a confidence interval for a population proportion p p is influenced by the standard error, which is calculated as SE=p^(1−p^)n SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} , where n n is the sample size
step 2
When p^(1−p^) \hat{p}(1-\hat{p}) is larger, the standard error increases, leading to a wider confidence interval. Conversely, when p^(1−p^) \hat{p}(1-\hat{p}) is smaller, the standard error decreases, resulting in a narrower confidence interval
step 3
Therefore, the statement that a confidence interval will be narrower when p^(1−p^) \hat{p}(1-\hat{p}) is larger is incorrect, confirming that the answer choice "False" is indeed correct
Answer
B
Key Concept
Confidence intervals for proportions are affected by the standard error, which depends on p^(1−p^) \hat{p}(1-\hat{p}) and sample size.
Explanation
A larger p^(1−p^) \hat{p}(1-\hat{p}) results in a wider confidence interval, while a smaller p^(1−p^) \hat{p}(1-\hat{p}) leads to a narrower interval.
A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of .996 ounces with a standard deviation of .004 ounces. How many pieces of candy must we sample if we want to be 99 percent confident that the sample mean is within .001 ounces of the true mean?
Solution by Steps
step 1
To determine the required sample size n n for a specified margin of error E E at a given confidence level, we use the formula: n=(Z⋅σE)2 n = \left( \frac{Z \cdot \sigma}{E} \right)^2 where Z Z is the Z-score corresponding to the desired confidence level, σ \sigma is the standard deviation, and E E is the margin of error
step 2
For a 99% confidence level, the Z-score is approximately Z=2.576 Z = 2.576 . The standard deviation σ \sigma is given as 0.004 0.004 ounces, and the margin of error E E is 0.001 0.001 ounces
step 3
Plugging in the values into the formula: n=(2.576⋅0.0040.001)2 n = \left( \frac{2.576 \cdot 0.004}{0.001} \right)^2
step 4
Calculating the expression: n=(0.0103040.001)2=(10.304)2≈106.19 n = \left( \frac{0.010304}{0.001} \right)^2 = (10.304)^2 \approx 106.19
step 5
Since the sample size must be a whole number, we round up to the nearest whole number: n=107 n = 107
Answer
107 pieces of candy must be sampled.
Key Concept
Sample size determination for estimating a population mean with a specified margin of error at a given confidence level.
Explanation
The calculation shows that to be 99% confident that the sample mean is within 0.001 ounces of the true mean, a sample size of 107 pieces of candy is required.
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